备注:在 LeetCode 中的运行时间也不是特别慢。 Java 实现 importjava.util.Map;importjava.util.HashMap;importjava.util.List;importjava.util.ArrayList;classSolution{publicList<Integer>topKFrequent(int[] nums,intk){// 统计元素的频率Map<Integer, Integer> freqMap =newHashMap<>();for(intnum : nums...
classSolution(object): deftopKFrequent(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ n=len(nums) cntDict=collections.defaultdict(int) foriinnums: cntDict[i]+=1 freqList=[[]foriinrange(n+1)] forpincntDict: freqList[cntDict[p]]+=p, ans=[...
Top K Frequent Elements - LeetCode https://leetcode.com/problems/top-k-frequent-elements/solution/ Python O(n) solution without sort, without heap, without quickselect - LeetCode Discuss https://leetcode.com/problems/top-k-frequent-elements/discuss/81697/Python-O(n)-solution-without-sort-with...
【leetcode】347. Top K Frequent Elements 题目地址:https://leetcode.com/problems/top-k-frequent-elements/ 从一个数组中求解出现次数最多的k个元素,本质是top k问题,用堆排序解决。 关于堆排序,其时间复杂度在最好和最坏的场景下都是O(nlogn)。 一开始想定义一个结构体,包含元素和元素个数两个成员,后...
You may assumekis always valid, 1 ≤k≤ number of unique elements. Your algorithm's time complexity must be better than O(nlogn), wherenis the array's size. 对于这种统计出现个数的题目,一般用map或unordered_map进行统计先,我开始的想法是通过统计之后,再用sort函数对pair的second进行排序输出,不过...
leetcode347 Top K Frequent Elements 1"""2Given a non-empty array of integers, return the k most frequent elements.3Example 1:4Input: nums = [1,1,1,2,2,3], k = 25Output: [1,2]6Example 2:7Input: nums = [1], k = 18Output: [1]9"""10"""11用dict实现的木桶排序12解法一:...
class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: # 统计 nums 中每个数字出现的次数, # 时间复杂度为 O(n) ,空间复杂度为 O(n) num_to_cnt = Counter(nums) #将 num_to_cnt 中的数字及其出现次数收集到数组中, # 时间复杂度为 O(n) ,空间复杂度为 O(n)...
You may assumekis always valid, 1 ≤k≤ number of unique elements. Your algorithm's time complexity must be better than O(nlogn), wherenis the array's size. 题目:给定一个非空数组,求出现次数最多的 k 个元素。 思路比较直接, 1. 将全部元素放到 Hashtable 里面,统计各个元素的出现次数 ...
3. 把top k Map.Entry入列 4. 把top k Map.Entry.getKey() 放入ArrayList 中 publicclassSolution {publicList<Integer> topKFrequent(int[] nums,intk) {//1. iterator all the distinct elements and countMap<Integer, Integer> map =newHashMap<Integer, Integer>();for(inti = 0; i < nums.lengt...
LeetCode - 347. Top K Frequent Elements Given a non-empty array of integers, return thekmost frequent elements. Example 1: Input: nums =[1,1,1,2,2,3], k =2 Output:[1,2] Example 2: Input: nums =[1], k =1 Output:[1]...